Reconstructing a signal using an equilizer filter function

[Evo8]

Reconstructing a signal using an equilizer "filter" function

Homework Statement

A low pass signal g(t) is sampled at a rate of $f_{s} > 2B$ needs to be reconstructed. Sampling interval is $T_s=\frac{1}{f_s}$

The reconstruction pulse to be used is

$$p(t)=∏(\frac{t}{T_s}-\frac{1}{2})$$

Specify the equalizer filter $E(f)$ to recover $g(t)$

Homework Equations

$$E(f)P(f)=0 / |f|>f_s-B$$
$$E(f)P(f)=T_s / |f|<B$$

$$E(f)=T_s \frac{ \pi f}{sin( \pi f T_p)} ≈ \frac{T_s}{T_p} \ \ |f|≤B$$

The Attempt at a Solution

There is one example in my text and it is not very clear at all.

I understand my "sampling interval" is $T_s=\frac{1}{f_s}$ What is $T_p$?

if my sampling frequency is greater then 2B then my last equation posted above does not apply right?

Im confused as where to start..

Thanks in advance for any help

 

[rude man]

Change $$ to ## for your reconstruction pulse? And is that the world's largest pi in front of it or does it mean "product"?

 

[Evo8]

Sorry about the code error i guess i was in a hurry. Trying to balance work and homework.

to answer your second question its both i guess? I never really have seen that notation before but now that you mention it, it makes more sense. The text I am using and my professor for that matter both use the capital pi and the lower case pi but not interchangeably.

 

[rude man]

Sorry about the code error i guess i was in a hurry. Trying to balance work and homework.

to answer your second question its both i guess? I never really have seen that notation before but now that you mention it, it makes more sense. The text I am using and my professor for that matter both use the capital pi and the lower case pi but not interchangeably.

Generally, upper-case pi denotes product and lower-case = 3.14...

Need to find out which was intended because this looks like a more challenging problem than the last one, even though it seems to use a similar sampling pulse train.

 

[Evo8]

Generally, upper-case pi denotes product and lower-case = 3.14...

Need to find out which was intended because this looks like a more challenging problem than the last one, even though it seems to use a similar sampling pulse train.

Ok so i finally found what the books notation means when using $∏$

Taken from the book:

Unit Rectangular Function
We use the pictorial notation $∏(x)$ for a rectangular pulse of unit height and unit width, centered at the origin as shown in Fig 3.6a

$∏(x)= \left\{ \begin{array}{ll} 1 & \mbox{|x|≤ 1/2}\\ 0.5 & \mbox{|x|=1/2}\\ 0 & \mbox{|x|>1/2}.\end{array} \right.$

Notice that the rectangular pulse in fig. 3.6b is the unit rectangular pulse $∏(x)$ expanded by a factor of $\tau$ and therefore can be expressed as $∏(x/ \tau)$. Observe that the denominator $\tau$ in $∏(x/ \tau)$ indicated the width of the pulse.

So I don't believe that it means the "product." that being said I am not sure I fully see how to apply this..

Sorry for the late response.

 

[rude man]

OK, I need to mull this one over. Like I said, more challenging than your last one.

However, it seems that the reconstruction pulse is identical to the one used in your last problem [sampled g(t)], except due to the "1/2" in the PI function it looks like it's not centered at t = 0 but starts at t = 0. Stay tuned ...

 

[berkeman]

Due to the advanced nature of thie problem, I'm going to grant rude_man some extra lattitude in providing help in this thread.

 

[rude man]

I was wrong when I said your reconstruction pulse is the same as the pulse-train version of g(t) in your last problem.

In this case what is happening is that the signal is sampled every T_s seconds via what's called a zero-order hold process. You get a sequence of FLAT pulses, each representing the signal at the sampling instants.

This is a pretty involved subject and even if I gave you hints you'd be hard-pressed to follow them I think. So instead I will direct you to a link that discusses the theory. Read the section starting with "Sampling with a zero-order hold" down to the "Interpolation" section.

What you need to understand in this writeup is:
1. The meaning of your reconstruction pulse definition is that x(t) is zero-order-hold sampled. The ZOH output is x_s(t).
2. The ZOH process is equivalent to delta-function sampling process followed by a filter with impulse response h₀(t). The output of the h₀(t) filter, which is the same as the output of the ZOH process, is inputed to a second filter h_Y(t).
3.h_Y(t) must be such that h₀(t) * h_Y(t) = the impulse response of an ideal low-pass filter, the frequency spectrum of which H(jw) = 1 from w = 0 to w= 2πB and = 0 for w > 2πB. This fact is the result of Dirac-delta sampling and I cannot go into the details of this - it's elementary but too lengthy. (The beginning of this link does that to some extent). The frequency-domain of this ideal low-pass filter is H(jω).
4. So the filter transfer function H_Y(jw) you are looking for is their equation 2.

The link is http://faraday.ee.emu.edu.tr/eeng420...s/lecture4.doc

 

[rude man]

I tried the link & it didn't come back.

If you send me you privatre e-mail I can e-mail you the file. If that's OK with Mr. Berkeman.

 

[Evo8]

I was able to download the lecture4.doc and pdf.

http://faraday.ee.emu.edu.tr/eeng420/
If you click on hte URL above and then navigate to lecture notes on the left hand side then scroll down to lecture notes and select lecture4.doc or .pdf i believe that is what you meant to link.I took another look through the book and did some more research online. I don't think this problem is meant to be so difficult..

According to my text (see screen shot below)

$E(f)=T_s/P(f)$ and $P(f)$ is the Fourier transform of my $p(t)$ and the i found the Fourier transform of a rectangular function is the following
$$∏(\frac{t}{\tau})= \tau sinc( \pi f \tau)$$

and
$$If\ \ g(t)\Leftrightarrow G(f)$$
$$Then \ \ g(t-t_0) \Leftrightarrow G(f) \exp^{-j2 \pi ft_0}$$

So I get $P(f)=T_{s} sinc( \pi f T_{s}) \exp^{-j \pi f}$

And if $E(f)=\frac{T_s}{P(f)} \ \ and\ T_s=\frac{1}{f_s}\ \ where\ f_s>2B$

Then I would get $E(f)=P(f)f_s$ which yields

$$E(f)=\frac{1}{2B} \ sinc( \pi f T_s) \exp^{-j \pi f}$$Maybe?...

However I feel that i handling the $f_s>2B$ incorrectly..

 

[rude man]

I was able to download the lecture4.doc and pdf.

http://faraday.ee.emu.edu.tr/eeng420/
If you click on hte URL above and then navigate to lecture notes on the left hand side then scroll down to lecture notes and select lecture4.doc or .pdf i believe that is what you meant to link.


I took another look through the book and did some more research online. I don't think this problem is meant to be so difficult..

According to my text (see screen shot below)

$E(f)=T_s/P(f)$ and $P(f)$ is the Fourier transform of my $p(t)$ and the i found the Fourier transform of a rectangular function is the following
$$∏(\frac{t}{\tau})= \tau sinc( \pi f \tau)$$

and
$$If\ \ g(t)\Leftrightarrow G(f)$$
$$Then \ \ g(t-t_0) \Leftrightarrow G(f) \exp^{-j2 \pi ft_0}$$

So I get $P(f)=T_{s} sinc( \pi f T_{s}) \exp^{-j \pi f}$

EDIT: sorry, I didn't realize that the textbook p(t) had T_p in it but your problem has T_s instead. So just replace T_p with T_s since for the problem, T_p = T_s which makes
p(t) = ∏(t/T_s - 1/2)

So your text has already defined E(f) in terms of P(f) with T_p = T_s and all you need to do is perform
E(f) = T_s/P(f), |f| <= B
= arbitrary, B < |f| < 1/T_s - B
= 0, |f| > 1/T_s - B.

So, you already have your P(f) which is the result of the zero-order hold process I discussed earlier, so this is just math. Since this filter is NOT REALIZABLE it's hard to know to what extent you're supposed to simplify the expression for E(f). I mean, in a way the answer is there already.

In real life no filter can have zero gain for |f| > 1/T_s - B or the ideal gain for |f| < B. All such E(f) are realizable approximations. The farther the spread between 2B and 1/T_s, the sloppier the filter can be, so in real life people sample at rates significantly greater than 1/T_s = 2B.

Example: a CD player samples at a rate of at least 44 KHz even though the realistic cutoff frequency is about 15 KHz so the Nyquist rate would be 30 KHz.

And if $E(f)=\frac{T_s}{P(f)} \ \ and\ T_s=\frac{1}{f_s}\ \ where\ f_s>2B$

Then I would get $E(f)=P(f)f_s$ ..

Now this you need to look at more carefully! (It would be nice if you could compute E(f) so easily! Hint: if y = a/x then y ≠ ax

Since you were able to open the faraday file link I still think perusing it would be beneficial. You are going to have to deal with inverting P(f) one way or another.

 

[Evo8]

Yeah my E(f) is wrong right there. Just incorrect algebra. Not sure what I was going.

Giving it another go I would get something like this $E(f)=\frac{T_s}{P(f)}=\frac{1}{P(f)f_s}$

If this is correct then my filter transfer function should look something like this
$$E(f)= \frac{1}{T_s \ sinc( \pi fT_s) \ e^{-j \pi f} (2B)}=\frac{1}{sinc(\frac{ \pi f}{2B} )e^{-j \pi f}}$$

And if I wanted to simplify further I would have to deal with inverting the P(f) as you mentioned. Or I would have attempted that before I "simplified" right?

 

[rude man]

Yeah my E(f) is wrong right there. Just incorrect algebra. Not sure what I was going.

Giving it another go I would get something like this $E(f)=\frac{T_s}{P(f)}=\frac{1}{P(f)f_s}$

If this is correct then my filter transfer function should look something like this
$$E(f)= \frac{1}{T_s \ sinc( \pi fT_s) \ e^{-j \pi f} (2B)}=\frac{1}{sinc(\frac{ \pi f}{2B} )e^{-j \pi f}}$$

And if I wanted to simplify further I would have to deal with inverting the P(f) as you mentioned. Or I would have attempted that before I "simplified" right?

Not quite. Look again at the book's definition of E(f) and substitute T_s for T_p since their formula for E(f) was developed with arbitrary pulse width T_p whereas your pulse width = T_s = 1/f_s.

You wind up with the 1/sinc()e^-j function but you also need to specify the behavior of E(f) for the region f > B since your problem said that f_s > 2B. Had it said f_s = 2B there would not be a "flexible" region. The E(f) filter would have to pass everything for 0 < f < B and reject everything above f = B. If we sample faster than 2B we can allow "slop" in E(f) which is the "flexible" region. Then, for f > f_s - B the filter E(f) needs to reject all signals since that is where the first "aliased" spectrum starts.

In the link I sent you their eq. 2 is identical with your E(f). Their H(jw) is the ideal low-pass filter which is H(jw) = 1, f < B and H(jw) = 0, f > B. That paper nicely shows how the aliased specrum develops as f_s varies above and below 2B.

As I said, this E(f) is not realizable. Practical filters can never meet these E(f) specifications. So what is done in practice is to increase f_s well above 2B in order to allow a wider "flexible" region and an acceptably low gain for when the first aliased spectrum intrudes, at f = f_s - B.